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Unformatted text preview: IOE 316 Fall 2007 – Homework 2 Solutions Due November 6, 2007 1. (15 points) Consider the following Markov chain: (a) (5 points) Find the values of x and y . (b) (10 points) Find 2step transition probability matrix P 2 SOLUTION: (a) (5 points) To find x , note that x must satisfy x + P 21 + P 22 = 1. Since P 21 = 0 . 25 and P 22 = 0 . 5, x = 0 . 25. Likewise, y must satisfy P 00 + y + P 02 = 1. WIth P 00 = 0 . 5 and P 02 = 0, we find that y = 0 . 5. (b) (10 points) With x and y known, the transition probability matrix P can be written as follows. P = . 5 . 5 . 5 . 5 . 25 0 . 25 0 . 5 To find P 2 , simply calculate P 2 = P · P = . 5 . 5 . 5 . 5 . 375 0 . 375 0 . 25 1 2. (25 points) A Markov chain { X n , n ≥ } with states 0 , 1 , 2, has the transition probability matrix P = 1 2 1 6 1 3 2 3 1 3 1 2 1 2 If P { X = 0 } = P { X = 1 } = 1 4 , find E [ X 3 ]. SOLUTION: (25 points) E [ X 3 ] = ∑ 2 i =0 iP ( X 3 = i ) = 0 · P ( X 3 = 0) + 1 · P ( X 3 = 1) + 2 · P ( X 3 = 2) To find P ( X 3 = 0) we use the Law of Total Probability P ( X 3 = 0) = P ( X 3 = 0  X = 0) P ( X = 0)+ P ( X 3 = 0  X = 1) P ( X = 1)+ P ( X 3 = 0  X = 2) P ( X = 2) = P 3 00 P ( X = 0) + P 3 10 P ( X = 1) + P 3 20 P ( X = 2) Similarly we can find P ( X 3 = 1) and P ( X 3 = 2)....
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 Spring '08
 Dolinskaya
 Probability theory, Markov chain, Andrey Markov, transition probability matrix, IOE

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